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    (Original post by physicsmaths)
    TRY this, draw that circle, the interior including boundary is the points that satisfy it.
    Now draw line x+y=k
    is it b?
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    (Original post by Mystery.)
    Thank you, did you just try coordinates to see? like (-1,0) (1,0)

    Also is the answer A to this :
    Well I thought more about how for any given x you can have either the positive or negative value of y, so the x axis must be a line of symmetry.

    For the question i'm quoting now, I personally thought of it as a circle, and ended up with b doing it like that.
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    (Original post by physicsmaths)
    People get in with really low scores but no way its the average.
    Check this official info out.
    https://www.whatdotheyknow.com/reque...matics_degrees
    So people who sat the 2014 paper for 2015 entry.
    Average score of all Imperial applicants:51
    Average score of successful: 66
    So nkt 20 marks lower.
    My friend got a 3A* offer this year for getting 48 in mat.


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    wait so doesn't that make the average score of Imperial applicants higher than Oxford ones? trippy
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    (Original post by CaiusMartius)
    wait so doesn't that make the average score of Imperial applicants higher than Oxford ones? trippy
    Not perhaps that surprising. You can't apply to Oxford and Cambridge both, so there will be a lot of Imperial applicants who are Cambridge, rather than Oxford applicants. Also Imperial don't interview - so it's perfectly possible to get an offer from Oxford with a moderate MAT score and good interviews. Imperial are more reliant on the test.

    PS Also looking now at the data, the average across all Imperial applicants was higher than Oxford's but the average amongst successful applicants was quite a bit lower.
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    Would I get any marks for 2015 Q2 (v) for this?
    To be a cube number, k^3 + 2k^2 + 2k + 1 = (k + a)^3
    Clearly a = 1 and hence for k^3 + 2k^2 + 2k + 1 to be a cubed number, it must be equal to (k + 1)^3

    However, (k + 1)^3 = k^3 + 3k^2 + 3k + 1 \neq k^3 + 2k^2 + 2k + 1 and hence k^3 + 2k^2 + 2k + 1 is not a cube number.

    Also, can someone explain why the same Q2 part (iv) uses 3^{2015} - 2^{2015} = (3^5 - 2^5)(\dots) when we established in (i) that a^{n + 1} - b^{n + 1} = (a - b)(\dots) - n = 2014?

    Thanks
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    (Original post by some-student)
    Would I get any marks for 2015 Q2 (v) for this?
    To be a cube number, k^3 + 2k^2 + 2k + 1 = (k + a)^3
    Clearly a = 1 and hence for k^3 + 2k^2 + 2k + 1 to be a cubed number, it must be equal to (k + 1)^3

    However, (k + 1)^3 = k^3 + 3k^2 + 3k + 1 \neq k^3 + 2k^2 + 2k + 1 and hence k^3 + 2k^2 + 2k + 1 is not a cube number.

    Also, can someone explain why the same Q2 part (iv) uses 3^{2015} - 2^{2015} = (3^5 - 2^5)(\dots) when we established in (i) that a^{n + 1} - b^{n + 1} = (a - b)(\dots) - n = 2014?

    Thanks
    Seems fine for full marks for part V, maybe state that it's because the expansion ends in +1?

    And there's no reason not to use the expansion of part i, so they did.
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    (Original post by RichE)
    Not perhaps that surprising. You can't apply to Oxford and Cambridge both, so there will be a lot of Imperial applicants who are Cambridge, rather than Oxford applicants. Also Imperial don't interview - so it's perfectly possible to get an offer from Oxford with a moderate MAT score and good interviews. Imperial are more reliant on the test.

    PS Also looking now at the data, the average across all Imperial applicants was higher than Oxford's but the average amongst successful applicants was quite a bit lower.
    Yup, that was me last year.
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    (Original post by RuairiMorrissey)
    Seems fine for full marks for part V, maybe state that it's because the expansion ends in +1?

    And there's no reason not to use the expansion of part i, so they did.
    (Original post by RuairiMorrissey)
    Seems fine for full marks for part V, maybe state that it's because the expansion ends in +1?

    And there's no reason not to use the expansion of part i, so they did.
    Thanks
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    (Original post by Mystery.)
    is the answer a ?
    What was the answer to this?
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    (Original post by Nonamebzja)
    What was the answer to this?
    There is no mark scheme.
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    Anyone know the answer to this ?
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    (Original post by Mystery.)
    There is no mark scheme.
    Oh
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    (Original post by Mystery.)
    Anyone know the answer to this ?
    X and y look equal
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    (Original post by Mystery.)
    Anyone know the answer to this ?
    I'm assuming only one of them is correct? In which case x = y as angles subtended from the same arc are equal, therefore x+y=2x
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    (Original post by Mystery.)
    There is no mark scheme.
    I think its b
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    (Original post by Mystery.)
    Anyone know the answer to this ?
    (this is probably wrong because other people got a different answer, oh well)
    I extended AP and BQ out of the circle until they touched (lets say at point 'S') and formed a kite shape. From there i said that angle RPS= pi - x (because angles on a straight line = pi) and from that i said that angle RQS= pi - (pi - x)= x (opposite angles in a kite add to pi). Finally, since RQB and RQS are on a straight line, they equal pi too, Therefore my answer is 'x + y = pi'.

    Can someone please check if this is wrong or right please? I feel like ive definitely made a mistake here
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    (Original post by AGFilsell)
    (this is probably wrong because other people got a different answer, oh well)
    I extended AP and BQ out of the circle until they touched (lets say at point 'S' and formed a kite shape. From there i said that angle RPS= pi - x (because angles on a straight line = pi) and from that i said that angle RQS= pi - (pi - x)= x (opposite angles in a kite add to pi). Finally, since RQB and RQS are on a straight line, they equal pi too, Therefore my answer is 'x + y = pi'.

    Can someone please check if this is wrong or right please? I feel like ive definitely made a mistake here
    That's what I did too.
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    Can anyone go through the 2005 multiple choice answers with me please?

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    Could someone explain q2 part IV of the 2015 paper? I dont understand how you can factorise it the way the mark scheme has
    cheersss
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    (Original post by theaverage)
    Can anyone go through the 2005 multiple choice answers with me please?

    Posted from TSR Mobile
    Yeah I am just doing it, what did you get?
 
 
 
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